Is ${90811}$ divisible by $3$ ?
A number is divisible by $3$ if the sum of its digits is divisible by $3$ . [ Why? First, we can break the number up by place value: $ \begin{eqnarray} {90811}= &&{9}\cdot10000+ \\&&{0}\cdot1000+ \\&&{8}\cdot100+ \\&&{1}\cdot10+ \\&&{1}\cdot1 \end{eqnarray} $ Next, we can rewrite each of the place values as $1$ plus a bunch of $9$ s: $ \begin{eqnarray} {90811}= &&{9}(9999+1)+ \\&&{0}(999+1)+ \\&&{8}(99+1)+ \\&&{1}(9+1)+ \\&&{1} \end{eqnarray} $ Now if we distribute and rearrange, we get this: $ \begin{eqnarray} {90811}= &&\gray{9\cdot9999}+ \\&&\gray{0\cdot999}+ \\&&\gray{8\cdot99}+ \\&&\gray{1\cdot9}+ \\&& {9}+{0}+{8}+{1}+{1} \end{eqnarray} $ Any number consisting only of $9$ s is a multiple of $3$ , so the first four terms must all be multiples of $3$ That means that to figure out whether the original number is divisible by $3 $ , all we need to do is add up the digits and see if the sum is divisible by $3$ . In other words, ${90811}$ is divisible by $3$ if ${ 9}+{0}+{8}+{1}+{1}$ is divisible by $3$ Add the digits of ${90811}$ $ {9}+{0}+{8}+{1}+{1} = {19} $ If ${19}$ is divisible by $3$ , then ${90811}$ must also be divisible by $3$ ${19}$ is not divisible by $3$, therefore ${90811}$ must not be divisible by $3$.